1999 OIM Problems/Problem 1

Problem

Find all positive integers that are less than 1000 and satisfy the following condition: the cube of the sum of their digits is equal to the square of that integer.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Insight: Every number that satisfies this must be a cube itself Proof/reasoning: let the sum of digits be $r$ and the original number be $n$ . Then $r^3 = n^2$ . If $n$ weren’t a cube, neither would $n^2$ , but it is. Therefore, $n$ is a cube.

Now we list out all cubes that are smaller than $1000$ $1,8,27,64,125,256,343,512,729$ $1^3 = 1^2 , 8^3 \neq 8^2, 9^3 = 27^2, 10^3 \neq 64^2, 8^3 \neq 125^2, 13^3 \neq 256^2 , 10^3 \neq 343^2 , 8^3 \neq 512^2 ,$ and $18^2 \neq 729^2$ . So the only integers that satisfy this condition are $1$ and $27$

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1999 OIM Problems/Problem 1

Problem

Solution

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